1.

The 4th term of an AP is OK. Prove that the 25th term of the AP is 3 times its 11th term.

Answer» We have given that4th\xa0term of an A.P.= a4\xa0= 0∴ a + (4 – 1)d = 0∴ a + 3d = 0∴ a = –3d ….(1)25th\xa0term of an A.P. = a25= a + (25 – 1)d= –3d + 24d ….[From the equation (1)]= 21d3 times 11th\xa0term of an A.P. = 3a11= 3[a + (11 – 1)d]= 3[a + 10d]= 3[–3d + 10d]= 3 × 7d= 21d∴ a25\xa0= 3a11i.e., the 25th\xa0term of the A.P. is three times its 11th\xa0term.


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